has a solution satisfying the initial condition , then it must satisfy the following integral equation:

Now we will solve this equation by the method of successive approximations.

Define as:

And define as

We will now prove that:

- If is bounded and the Lipschitz condition is satisfied, then the sequence of functions converges to a continuous function
- This function satisfies the differential equation
- This is the unique solution to this differential equation with the given initial condition.

## ProofEdit

First, we prove that lies in the box, meaning that . We prove this by induction. First, it is obvious that . Now suppose that . Then so that

. This proves the case when , and the case when is proven similarily.

We will now prove by induction that . First, it is obvious that . Now suppose that it is true up to n-1. Then

due to the Lipschitz condition.

Now,

.

Therefore, the series of series is absolutely and uniformly convergent for because it is less than the exponential function.

Therefore, the limit function exists and is a continuous function for .

Now we will prove that this limit function satisfies the differential equation.